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3 years ago

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14. Con base en los números complejos expresados en su forma trigonométrica:

1 answer:

Lorico [155]3 years ago

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Dino nuggets

Bro stop scrolling

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Write an equation in slope-intercept form of<br> the line that passes through (-6,4) and<br> (-1,2).

xz_007 [3.2K]

Answer:

Step-by-step explanation:

(-6 , 4) & (-1 , 2)

Slope = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$= \frac{2-4}{-1-[-6]}\\\\= \frac{-2}{-1+6}\\\\= \frac{-2}{5}$

m = -2/5 & (-6 , 4)

y -y1 = m(x -x1)

$y - 4 = \frac{-2}{5}(x - [-6])\\\\y - 4 = \frac{-2}{5}(x + 6)\\\\y - 4 = \frac{-2}{5}x + 6*\frac{-2}{5}\\\\y = \frac{-2}{5}x -\frac{12}{5}+4\\\\y=\frac{-2}{5}x-\frac{12}{5}+\frac{4*5}{1*5}\\\\y=\frac{-2}{5}x-\frac{12}{5}+\frac{20}{5}\\\\y=\frac{-2}{5}x+\frac{8}{5}$

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3 years ago

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SashulF [63]

Answer:

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Step-by-step explanation:

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3 years ago

Probabilities with possible states of nature: s1, s2, and s3. Suppose that you are given a decision situation with three possibl

amm1812

Answer:

1. $P(s_1|I)=\frac{1}{11}$

2. $P(s_2|I)=\frac{8}{11}$

3. $P(s_3|I)=\frac{2}{11}$

Step-by-step explanation:

Given information:

$P(s_1)=0.1, P(s_2)=0.6, P(s_3)=0.3$

$P(I|s_1)=0.15,P(I|s_2)=0.2,P(I|s_3)=0.1$

(1)

We need to find the value of P(s₁|I).

$P(s_1|I)=\frac{P(I|s_1)P(s_1)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}$

$P(s_1|I)=\frac{(0.15)(0.1)}{(0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3)}$

$P(s_1|I)=\frac{0.015}{0.015+0.12+0.03}$

$P(s_1|I)=\frac{0.015}{0.165}$

$P(s_1|I)=\frac{1}{11}$

Therefore the value of P(s₁|I) is $\frac{1}{11}$ .

(2)

We need to find the value of P(s₂|I).

$P(s_2|I)=\frac{P(I|s_2)P(s_2)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}$

$P(s_2|I)=\frac{(0.2)(0.6)}{(0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3)}$

$P(s_2|I)=\frac{0.12}{0.015+0.12+0.03}$

$P(s_2|I)=\frac{0.12}{0.165}$

$P(s_2|I)=\frac{8}{11}$

Therefore the value of P(s₂|I) is $\frac{8}{11}$ .

(3)

We need to find the value of P(s₃|I).

$P(s_3|I)=\frac{P(I|s_3)P(s_3)}{P(I|s_1)P(s_1)+P(I|s_2)P(s_2)+P(I|s_3)P(s_3)}$

$P(s_3|I)=\frac{(0.1)(0.3)}{(0.15)(0.1)+(0.2)(0.6)+(0.1)(0.3)}$

$P(s_3|I)=\frac{0.03}{0.015+0.12+0.03}$

$P(s_3|I)=\frac{0.03}{0.165}$

$P(s_3|I)=\frac{2}{11}$

Therefore the value of P(s₃|I) is $\frac{2}{11}$ .

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